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\title{微分方程数值解\ 第8次作业}
\author{李之琪 22235056}

\begin{document}
\maketitle
\section*{3.3.1}
当以下条件满足时，能量保持不变：(i)$A$与单位矩阵$I$相似，即存在矩阵
$S$,使得$S^{-1}AS =
I$；(ii)$S^{-1}BS$是个对角矩阵，且其所有非零元是纯虚数。下面给出说明：

对原方程作Fourier变换得到
\begin{eqnarray}
  \begin{aligned}
    \hat{u_t} = i\omega A\hat{u} + B\hat{u}.
  \end{aligned}
\end{eqnarray}
令$\tilde{u} = S^{-1}\hat{u}$，则有
\begin{eqnarray}
  \begin{aligned}
    \tilde{u_t} = i\omega I\tilde{u} + \tilde{B}\tilde{u},
  \end{aligned}
\end{eqnarray}
这里$\tilde{B} = S^{-1}BS$，根据条件，$\tilde{B}$是对角矩阵且所有非零
元是纯虚数。以上方程的解为
\begin{eqnarray}
  \begin{aligned}
    \tilde{u} = e^{(i\omega I + \tilde{B})t}\tilde{f}(\omega),
  \end{aligned}
\end{eqnarray}
注意到$i\omega I + \tilde{B}$仍是个所有非零元是纯虚数的对角阵，故
\begin{eqnarray}
  \begin{aligned}
    \left\|u(\cdot,t) \right\|^2 &= \sum_{\omega}\left|Se^{(i\omega I +
        \tilde{B})t}S^{-1}\tilde{f}(\omega)\right|^2\\
    &= \sum_{\omega}\left|\tilde{f}(\omega)\right|^2 = \left\|u(\cdot,0) \right\|^2.
  \end{aligned}
\end{eqnarray}
\section*{3.4.1}
记
\begin{eqnarray}
  \begin{aligned}
    E(t) = \left\|u(\cdot,t) \right\|^2+2\delta\int_{0}^t\left\|u_x(\cdot,\xi) \right\|^2,
  \end{aligned}
\end{eqnarray}
这里$\delta$来自parabolic条件，即$Re\lambda(A) \ge \delta > 0$。下面我
们略去记号$(\cdot,t)$，则有
\begin{eqnarray}
  \begin{aligned}
    \frac{\text{d} E(t)}{\text{d} t} &= (u,u_t) + (u_t,u) +2\delta
    (u_x,u_x)\\
    &= (u,Au_{xx}) + (Au_{xx},u) +2\delta
    (u_x,u_x)\\
    &= (A + A^{\ast})(u,u_{xx})  +2\delta
    (u_x,u_x)\\
    &\le 2\delta[(u,u_{xx}) + (u_x,u_x)] = 0.
  \end{aligned}
\end{eqnarray}
故而
\begin{eqnarray}
  \begin{aligned}
    \left\|u(\cdot,t)
    \right\|^2+2\delta\int_{0}^t\left\|u_x(\cdot,\xi) \right\|^2 =
    E(t) \le E(0) = \left\|u(\cdot,0)
    \right\|^2.
  \end{aligned}
\end{eqnarray}
\section*{3.4.2}
结论不成立。不妨考虑一维情形
\begin{eqnarray}
  \begin{aligned}
    u_t = \delta u_{xx} + u_x.
  \end{aligned}
\end{eqnarray}
则
\begin{eqnarray}
  \begin{aligned}
    \frac{\text{d} E(t)}{\text{d} t} &= (u,u_t) + (u_t,u) +2\delta
    (u_x,u_x)\\
    &= 2\delta (u, u_{xx}) + 2 (u,u_{x}) +2\delta
    (u_x,u_x).
  \end{aligned}
\end{eqnarray}
由于我们总是能取适当的初值使得$(u(\cdot,0),u_{x}(\cdot,0)) > 0$，故
$\dfrac{\text{d} E(0)}{\text{d} t} > 0$,从而对充分小的$t$，一定有
\begin{eqnarray}
  \begin{aligned}
    \left\|u(\cdot,t)
    \right\|^2+2\delta\int_{0}^t\left\|u_x(\cdot,\xi) \right\|^2 =
    E(t) > E(0) = \left\|u(\cdot,0)
    \right\|^2.
  \end{aligned}
\end{eqnarray}
即结论不成立。
\section*{3.5.1}
题中所述的情形是不可能的。对方程$u_t = Au_x$，Petrovskii条件为
\begin{eqnarray}
  \begin{aligned}
    Re\lambda(i\omega A) < \alpha,
  \end{aligned}
\end{eqnarray}
这里$Re\lambda$表示矩阵的特征值的实部。若题中所述条件成立，一方面，存在$\alpha > 0$，对任意$\omega$，有
$Re\lambda(i\omega A) < \alpha$；另一方面，由于$Re\lambda(i\omega A) < 0$不恒成立，
故一定存在$i,\omega_0,\delta > 0$，使得$Re\lambda_i(i\omega_0 A) \ge
\delta$。再结合$\omega$的任意性，对充分大的$\omega = k\omega_0$(取$k =
2\alpha/ \delta$即可)，一定
有$Re\lambda_i(i\omega A) > \alpha$，矛盾。故题中所述条件不可能成
立。
\end{document}

